Problem: Simplify and expand the following expression: $ \dfrac{2}{2r + 20}- \dfrac{4}{r - 6}- \dfrac{r}{r^2 + 4r - 60} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the first term: $ \dfrac{2}{2r + 20} = \dfrac{2}{2(r + 10)}$ We can factor the quadratic in the third term: $ \dfrac{r}{r^2 + 4r - 60} = \dfrac{r}{(r + 10)(r - 6)}$ Now we have: $ \dfrac{2}{2(r + 10)}- \dfrac{4}{r - 6}- \dfrac{r}{(r + 10)(r - 6)} $ The least common multiple of the denominators is: $ 2(r + 10)(r - 6)$ In order to get the first term over $2(r + 10)(r - 6)$ , multiply by $\dfrac{r - 6}{r - 6}$ $ \dfrac{2}{2(r + 10)} \times \dfrac{r - 6}{r - 6} = \dfrac{2(r - 6)}{2(r + 10)(r - 6)} $ In order to get the second term over $2(r + 10)(r - 6)$ , multiply by $\dfrac{2(r + 10)}{2(r + 10)}$ $ \dfrac{4}{r - 6} \times \dfrac{2(r + 10)}{2(r + 10)} = \dfrac{8(r + 10)}{2(r + 10)(r - 6)} $ In order to get the third term over $2(r + 10)(r - 6)$ , multiply by $\dfrac{2}{2}$ $ \dfrac{r}{(r + 10)(r - 6)} \times \dfrac{2}{2} = \dfrac{2r}{2(r + 10)(r - 6)} $ Now we have: $ \dfrac{2(r - 6)}{2(r + 10)(r - 6)} - \dfrac{8(r + 10)}{2(r + 10)(r - 6)} - \dfrac{2r}{2(r + 10)(r - 6)} $ $ = \dfrac{ 2(r - 6) - 8(r + 10) - 2r} {2(r + 10)(r - 6)} $ Expand: $ = \dfrac{2r - 12 - 8r - 80 - 2r}{2r^2 + 8r - 120} $ $ = \dfrac{-8r - 92}{2r^2 + 8r - 120}$ Simplify: $ = \dfrac{-4r - 46}{r^2 + 4r - 60}$